The Real Line — Worked Problems

What this book adds

The free chapters on this topic teach the ideas and the traps. This is the companion problem set: eleven questions at the length and difficulty CSIR NET actually sets, each solved completely rather than sketched — followed by a compressed reference sheet and six problems for you to check yourself against, answers only.

Work them in order. Several later solutions reuse a result from an earlier one, the way an exam paper itself rewards having done the easy part first.

Quick reference

Supremum. $s = \sup A$ iff $s$ is an upper bound and, for every $\varepsilon > 0$, some $a \in A$ has $a > s - \varepsilon$. Not necessarily attained.

Archimedean property. For every $x \in \mathbb{R}$ there is $n \in \mathbb{N}$ with $n > x$.

The implication chain. convergent $\Rightarrow$ Cauchy $\Rightarrow$ bounded $\Rightarrow$ has a convergent subsequence. Only Cauchy $\Rightarrow$ convergent needs completeness; none of the arrows reverse without it.

Monotone convergence. Increasing and bounded above $\Rightarrow$ converges, to the supremum. Prove monotone, prove bounded, then solve $L = f(L)$ — never before.

limsup / liminf. $\limsup a_n = \lim_{n} \sup_{k \ge n} a_k$, the largest subsequential limit. $\liminf a_n \le \limsup a_n$ always; equality (and finite) iff the sequence converges. Subadditive only: $\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n$, with equality restored the moment one of the two sequences converges.

Choosing a series test, in order: $a_n \not\to 0 \Rightarrow$ diverges, free. Shape $1/n^p$ or geometric $\Rightarrow$ comparison ($\sum 1/n^p$ converges iff $p>1$). Factorials or $n^n$ $\Rightarrow$ ratio. $n$-th powers $\Rightarrow$ root (strictly stronger than ratio). Ratio $\to 1$ $\Rightarrow$ Raabe: $R = \lim n(a_n/a_{n+1} - 1)$, converges for $R>1$, diverges for $R<1$. Logs, terms decreasing $\Rightarrow$ Cauchy condensation: $\sum a_n$ and $\sum 2^k a_{2^k}$ share a fate. Alternating $\Rightarrow$ Leibniz, then check $\sum |a_n|$ separately for absolute convergence.

Worked problems

1. A nested family of open intervals whose intersection is closed

Problem. For $n \in \mathbb{N}$, let $A_n = \left(-\dfrac{1}{n},\ 1 + \dfrac{1}{n}\right)$. Find $\bigcup_{n=1}^{\infty} A_n$ and $\bigcap_{n=1}^{\infty} A_n$.

Solution.

The family is nested and decreasing: $-1/n$ increases toward $0$ and $1+1/n$ decreases toward $1$ as $n$ grows, so $A_1 \supseteq A_2 \supseteq A_3 \supseteq \cdots$.

For a decreasing nested family the union is simply the first, largest set:

$\bigcup_{n=1}^{\infty} A_n = A_1 = (-1, 2).$

The intersection needs more care. First, $[0,1] \subseteq \bigcap_n A_n$: for every $n$, $0 > -1/n$ and $1 < 1+1/n$, so both endpoints of $[0,1]$ lie in every $A_n$, and so does everything between them.

Now take any $x < 0$, say $x = -\varepsilon$ with $\varepsilon > 0$.

By the Archimedean property there is $n$ with $n > 1/\varepsilon$, i.e. $1/n < \varepsilon$, so $-1/n > -\varepsilon = x$ and $x \notin A_n$.

So no point below $0$ survives every $A_n$; the symmetric argument disposes of every $x > 1$. Hence

$\bigcap_{n=1}^{\infty} A_n = [0, 1].$

Every $A_n$ is open, and their intersection is closed. This is not a contradiction — only a finite intersection of open sets is guaranteed open; an infinite one need not be, and this is the standard example examiners reach for when a statement quietly drops the word "finite."

2. A supremum that is attained and an infimum that is not, in the same set

Problem. Let $A = \{(-1)^n + \tfrac{1}{n} : n \in \mathbb{N}\}$. Find $\sup A$ and $\inf A$, and say whether each is attained.

Solution.

Split on parity.

For even $n$: $(-1)^n + 1/n = 1 + 1/n$, which is largest at the smallest even $n$, namely $n=2$, giving $3/2$; it decreases toward $1$ (never reaching it) as $n$ grows.

For odd $n$: $(-1)^n + 1/n = -1 + 1/n$, largest at $n=1$, giving $0$; it decreases toward $-1$ (never reaching it) as $n$ grows.

Comparing the two tracks: the largest value overall is $3/2$, attained at $n=2$, so $\sup A = 3/2$, attained.

The smallest values come from the odd track, which descends toward $-1$ without ever reaching it (equality would need $1/n = 0$), so $\inf A = -1$, not attained.

The point of pairing these in one set: there is no rule that a set's sup and inf are either both attained or both not — each is decided separately, by its own track.

3. Density of $\mathbb{Q}$ in $\mathbb{R}$, from the Archimedean property

Problem. Prove that for any reals $a < b$, there is a rational number $r$ with $a < r < b$.

Solution.

Since $b - a > 0$, the Archimedean property gives $n \in \mathbb{N}$ with $n > \dfrac{1}{b-a}$, i.e. $\dfrac{1}{n} < b - a$.

Let $m$ be the smallest integer with $m > na$. Such an $m$ exists: apply the Archimedean property to $na$ to get some integer above it, then take the least one — a non-empty set of integers bounded below has a least element.

Minimality of $m$ gives $m - 1 \le na$, i.e. $m \le na + 1$.

Now put $r = m/n$.

From $m > na$: $r = m/n > a$.

From $m \le na+1$: $r = m/n \le a + 1/n < a + (b-a) = b$.

So $a < r < b$ with $r = m/n \in \mathbb{Q}$ — the rationals are dense in $\mathbb{R}$ only because the reals fill every gap densely enough to make this construction land inside $(a,b)$ no matter how short it is.

4. A recursively defined sequence, proved monotone and bounded, limit found last

Problem. Define $a_1 = 2$ and $a_{n+1} = \dfrac{1}{2}\left(a_n + \dfrac{2}{a_n}\right)$ for $n \ge 1$. Show $(a_n)$ converges, and find its limit.

Solution.

First, $a_n \ge \sqrt2$ for every $n \ge 1$.

This is immediate from AM–GM: for any $x>0$, $\dfrac{1}{2}(x + 2/x) \ge \sqrt{x \cdot 2/x} = \sqrt2$, so $a_{n+1} \ge \sqrt2$ for $n \ge 1$, and $a_1 = 2 \ge \sqrt2$ as well.

Next, $(a_n)$ is decreasing from $n=1$ onward:

$a_{n+1} - a_n = \frac{1}{2}\left(\frac{2}{a_n} - a_n\right) = \frac{2 - a_n^2}{2a_n}.$

Since $a_n \ge \sqrt2$, $a_n^2 \ge 2$, so the numerator is $\le 0$ and $a_{n+1} \le a_n$.

So $(a_n)$ is decreasing and bounded below by $\sqrt2$: by the monotone convergence theorem it converges. Call the limit $L$.

Only now, with existence established, take the limit of both sides of the recursion:

$L = \frac{1}{2}\left(L + \frac{2}{L}\right) \implies 2L^2 = L^2 + 2 \implies L^2 = 2.$

Since every $a_n \ge \sqrt2 > 0$, $L \ge \sqrt2 > 0$, so $L = \sqrt2$.

This is Newton's method for $\sqrt2$ starting from $a_1=2$; it is also why the sequence converges so fast — $a_4$ already agrees with $\sqrt2$ to five decimal places.

5. Proving a sequence is Cauchy directly, without assuming a limit exists

Problem. Let $s_n = \displaystyle\sum_{k=1}^{n} \frac{1}{k^2}$. Prove $(s_n)$ is Cauchy.

Solution.

Take $m > n$. Then

$|s_m - s_n| = \sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} \frac{1}{k(k-1)} = \sum_{k=n+1}^{m} \left(\frac{1}{k-1} - \frac{1}{k}\right) = \frac{1}{n} - \frac{1}{m} < \frac{1}{n},$

the middle sum telescoping to just its first and last pieces, and the inequality $1/k^2 < 1/(k(k-1))$ holding for every $k \ge 2$.

Given $\varepsilon > 0$, the Archimedean property gives $N$ with $1/N < \varepsilon$.

Then for all $m > n \ge N$: $|s_m - s_n| < 1/n \le 1/N < \varepsilon$. So $(s_n)$ is Cauchy.

This is where completeness earns its keep — Cauchy in $\mathbb{R}$ means convergent, with no need to name the limit. (It happens to be $\pi^2/6$, which is a separate, much harder fact.)

The technique — bound the tail by a telescoping sum — is the standard route to proving a sequence of partial sums is Cauchy directly, and it works whenever the terms are eventually dominated by $1/(k(k-1))$ or $1/(k(k+1))$.

6. One sequence, four questions and both limits

Problem. Let $a_n = \dfrac{(-1)^n n}{n+1}$. Determine whether $(a_n)$ is (i) bounded, (ii) Cauchy, (iii) convergent, (iv) has a convergent subsequence — and compute $\limsup a_n$ and $\liminf a_n$.

Solution.

Split on parity again, since $(-1)^n$ is exactly the kind of factor that makes this the right move.

Even $n$: $a_n = \dfrac{n}{n+1} \to 1$. Odd $n$: $a_n = -\dfrac{n}{n+1} \to -1$.

(i) Bounded. $|a_n| = n/(n+1) < 1$ for every $n$, so yes.

(iv) Convergent subsequence. Yes — in fact two: the even-indexed terms converge to $1$ and the odd-indexed terms converge to $-1$.

(iii) Convergent. No. A sequence with two distinct subsequential limits cannot converge — if it did, every subsequence would converge to that same limit.

(ii) Cauchy. No, for the same reason: in $\mathbb{R}$, Cauchy and convergent coincide, and this sequence is not convergent.

limsup and liminf. The subsequential limits are exactly $1$ and $-1$, and there are no others, so $\limsup a_n = 1$ and $\liminf a_n = -1$ — consistent with $\liminf < \limsup$, which must hold given the sequence does not converge.

This is the shape almost every "decide everything about this sequence" question takes: find the tracks, read every answer off them, and the limsup/liminf question is not a fifth calculation but the same two numbers already found in part (iv).

7. limsup and liminf from residue-class tracks

Problem. Compute $\limsup a_n$ and $\liminf a_n$ for $a_n = \cos\!\left(\dfrac{n\pi}{3}\right) + \dfrac{1}{n}$.

Solution.

$\cos(n\pi/3)$ depends only on $n \bmod 6$, cycling through six values:

$n \equiv 0,1,2,3,4,5 \pmod 6 \ \longmapsto\ \cos\frac{n\pi}{3} = 1,\ \tfrac12,\ -\tfrac12,\ -1,\ -\tfrac12,\ \tfrac12.$

The added $1/n \to 0$ shifts every track down toward its limiting value without changing which values are largest and smallest.

Along $n \equiv 0 \pmod 6$: $a_n = 1 + 1/n \to 1$.

Along $n \equiv 3 \pmod 6$: $a_n = -1 + 1/n \to -1$.

No other residue class reaches outside $[-1/2, 1/2]$ in the limit, so these two are the extreme subsequential limits:

$\limsup a_n = 1, \qquad \liminf a_n = -1.$

This is the general method for any sequence built from a periodic piece plus a term tending to $0$: reduce to residue classes mod the period, take each track to its limit, and read off the largest and smallest.

8. When does subadditivity become equality?

Problem. Prove: if $a_n \to L$ (a genuine limit, not just bounded) and $(b_n)$ is any bounded sequence, then $\limsup(a_n + b_n) = L + \limsup b_n$.

Solution.

Let $M = \limsup b_n$ and fix $\varepsilon > 0$. Since $a_n \to L$, there is $N$ with $|a_n - L| < \varepsilon/2$ for all $n \ge N$.

Upper bound. For $n \ge N$:

$\sup_{k \ge n}(a_k+b_k) \le \sup_{k\ge n}\big((L+\varepsilon/2)+b_k\big) = L + \varepsilon/2 + \sup_{k \ge n} b_k.$

Letting $n \to \infty$ on both sides gives $\limsup(a_n+b_n) \le L + \varepsilon/2 + M$.

Lower bound. Symmetrically, $a_k > L - \varepsilon/2$ for $k \ge N$ gives $\sup_{k\ge n}(a_k+b_k) \ge L - \varepsilon/2 + \sup_{k\ge n} b_k$ for $n \ge N$, so $\limsup(a_n+b_n) \ge L - \varepsilon/2 + M$.

Both bounds hold for every $\varepsilon > 0$, so $\limsup(a_n+b_n) = L + M$.

Why this does not contradict $\limsup(a_n{+}b_n) \le \limsup a_n + \limsup b_n$ sometimes being strict, as in $(-1)^n$ against $(-1)^{n+1}$: that example has neither sequence converging.

The moment one side is forced to a genuine limit, it can no longer be out of phase with the other, and the inequality tightens to equality. Strictness is entirely a phase effect between two oscillating sequences — remove the oscillation from one side and it is gone.

9. Raabe's test, on a series the ratio test cannot settle

Problem. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(2n)!}{4^n (n!)^2}$ converges.

Solution.

Let $a_n = \dfrac{(2n)!}{4^n(n!)^2}$. The ratio:

$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{4^{n+1}((n+1)!)^2} \cdot \frac{4^n(n!)^2}{(2n)!} = \frac{(2n+1)(2n+2)}{4(n+1)^2} = \frac{2n+1}{2n+2} \longrightarrow 1.$

Inconclusive — exactly the signal to reach for Raabe. Its reciprocal:

$\frac{a_n}{a_{n+1}} = \frac{2n+2}{2n+1} = 1 + \frac{1}{2n+1}.$

So

$R = \lim_{n\to\infty} n\left(\frac{a_n}{a_{n+1}} - 1\right) = \lim_{n\to\infty} \frac{n}{2n+1} = \frac12.$

$R = 1/2 < 1$, so the series diverges.

This matches the asymptotic $a_n \sim 1/\sqrt{\pi n}$ from Stirling's formula — a $p=1/2$ comparison, well under the $p>1$ threshold — but Raabe reaches the same conclusion without needing Stirling at all, which is the actual point of having the test.

10. Cauchy condensation, worked in general rather than by example

Problem. For $n \ge 2$ and $p > 0$, decide for which $p$ the series $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p}$ converges.

Solution.

The terms $a_n = 1/(n(\log n)^p)$ are positive and decreasing, since both $n$ and $\log n$ increase — so Cauchy condensation applies: $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges.

Compute:

$2^k a_{2^k} = \frac{2^k}{2^k(\log 2^k)^p} = \frac{1}{(k \log 2)^p} = \frac{1}{(\log 2)^p}\cdot\frac{1}{k^p}.$

Dropping the constant $1/(\log2)^p$, this is a $p$-series in $k$, which converges iff $p>1$. So

$\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p} \text{ converges} \iff p > 1.$

In particular $\sum 1/(n\log n)$ ($p=1$) diverges and $\sum 1/(n\log^2 n)$ ($p=2$) converges — the specific pair the exam likes to set side by side is just this general result at two values of $p$, not two facts to memorise separately.

11. Absolute versus conditional, reusing Problem 10

Problem. Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log n}$ converges absolutely, converges conditionally, or diverges.

Solution.

Absolute values give $\sum 1/(n \log n)$, which is exactly Problem 10 at $p=1$ — divergent. So the series is not absolutely convergent.

For convergence itself, apply Leibniz to $a_n = 1/(n\log n)$: it is positive, decreasing for $n \ge 2$ (as argued in Problem 10), and $a_n \to 0$. So the alternating series converges.

Convergent, not absolutely convergent: conditionally convergent.

And because it is only conditionally convergent, Riemann's rearrangement theorem applies to it — reordering its terms can be made to sum to any real number at all, or to diverge, which is not true of Problem 5's series (that one converges absolutely, so every rearrangement sums to the same value).

Practice — check yourself

Answers only — the point is to test whether the methods above transfer, not to read another eleven solutions.

1. Find $\sup B$ and $\inf B$, and say whether each is attained, for $B = \{1 - 1/n : n \in \mathbb{N}\} \cup \{2\}$.
2. Is $a_n = (-1)^n\left(1 + \dfrac{1}{n^2}\right)$ Cauchy?
3. Compute $\limsup a_n$ and $\liminf a_n$ for $a_n = (-1)^n + \dfrac{1}{n}\sin\dfrac{n\pi}{2}$.
4. Does $\displaystyle\sum_{n=1}^\infty \frac{n!}{n^n}$ converge? (Ratio test.)
5. Is $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n}$ absolutely convergent, conditionally convergent, or divergent?
6. For $c_n = \sup\{1/k : k \ge n\}$, find $\lim_n c_n$.

Work them before reading on — the answers are right below.

Answers

1. $\sup B = 2$ (attained, it is the isolated point); $\inf B = 0$ (attained, at $n=1$).
2. No — the even and odd subsequences converge to $1$ and $-1$ respectively, so the sequence does not converge, hence is not Cauchy.
3. $\limsup a_n = 1$, $\liminf a_n = -1$ (the $\frac1n\sin(n\pi/2)$ term is squeezed to $0$ and does not move either extreme).
4. Converges: $a_{n+1}/a_n = (n/(n+1))^n \to 1/e < 1$.
5. Conditionally convergent: $\sum 1/\sqrt n$ is a $p=1/2$ series and diverges, but the alternating series itself converges by Leibniz.
6. $0$ — the tail supremum of a decreasing sequence is its first term, so $c_n = 1/n \to 0$.